QB. Find the area of the region bounded by the curves y = x², y = |2 - x²| and y = 2, which lies to the right of the line x = 1.
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QB. Find the area of the region bounded by the curves y = x², y = |2 - x²| and y
QB. Find the area of the region bounded by the curves y = x², y = |2 - x²| and y = 2, which lies to the right of the line x = 1.
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[tex]\large\underline{\sf{Solution-}}[/tex]
Given curves are
[tex]\rm :\longmapsto\:P_1 : y = {x}^{2} [/tex]
[tex]\rm :\longmapsto\:P_2 : y = |2 - {x}^{2} | [/tex]
[tex]\rm :\longmapsto\:L_1 : y = 2[/tex]
[tex]\rm :\longmapsto\:L_2 : x= 1[/tex]
Now,
[tex]\rm :\longmapsto\:P_1 : y = {x}^{2} [/tex]
represents the parabola whose vertex is (0, 0).
[tex]\rm :\longmapsto\:P_2 : y = |2 - {x}^{2} | [/tex]
represents the downward parabola whose vertex is at (0, 2) and intersects the x - axis at
[tex]\rm :\longmapsto\:( - \sqrt{2},0) \: and \: ( \sqrt{2},0)[/tex]
Now,
[tex]\begin{gathered}\begin{gathered}\bf\: |2 - {x}^{2} | = \begin{cases} &\sf{ {x}^{2} - 2 \: \: if \: x < - \sqrt{2} } \\ &\sf{2 - {x}^{2} \: if \: - \sqrt{2} \leqslant x \leqslant \sqrt{2} }\\ &\sf{ {x}^{2} - 2 \: if \: x > \sqrt{2} } \end{cases}\end{gathered}\end{gathered}[/tex]
[tex]\rm :\longmapsto\:L_1 : y = 2[/tex]
is a line parallel to x - axis passes through (0, 2).
Required area bounded between the curves with respect to x -axis is given by
[tex]\rm \: \: = \:\displaystyle\int_1^ {\sqrt{2}}(P_1 - P_2)dx \: + \: \displaystyle\int_{ \sqrt{2}} ^ 2(L_1 - P_2)dx[/tex]
[tex]\rm \: \: = \:\displaystyle\int_1^ {\sqrt{2}}( {x}^{2} -(2 - {x}^{2} ))dx \: + \: \displaystyle\int_{ \sqrt{2}} ^ 2(2 - ( {x}^{2} - 2) )dx[/tex]
[tex]\rm \: \: = \:\displaystyle\int_1^ {\sqrt{2}}( 2{x}^{2} -2 )dx \: + \: \displaystyle\int_{ \sqrt{2}} ^ 2(4 - {x}^{2})dx[/tex]
[tex]\rm \: \: = \: \bigg[\dfrac{ {2x}^{3} }{3} - 2x\bigg]_1^{ \sqrt{2}} + \bigg[\dfrac{ 4x - {x}^{3} }{3} \bigg]_{ \sqrt{2}}^{2}[/tex]
[tex]\rm \: \: = \:\dfrac{4 \sqrt{2} }{3} - 2 \sqrt{2} - \dfrac{2}{3} + 2 + 8 - \dfrac{8}{3} - 4 \sqrt{2} + \dfrac{2 \sqrt{2} }{3} [/tex]
[tex]\rm \: \: = \:\dfrac{20}{3} - 4 \sqrt{2} \: sq. \: units[/tex]
Formula Used :-
[tex]\rm :\longmapsto\:Area \: w.r.t \: x \: axis = \displaystyle\int_a^b \: y \: dx[/tex]
[tex]\rm :\longmapsto\:\displaystyle\int \: k \: dx = kx \: + \: c[/tex]
[tex]\rm :\longmapsto\:\displaystyle\int \: {x}^{n} \: dx = \dfrac{ {x}^{n + 1} }{n + 1} \: + \: c[/tex]